I was reading about factorials recently, and I happened to come across the curious, almost pseudo-Pythagorean-seeming fact that $$6!7!=10!$$ I was greatly intrigued by this, but couldn't think of any justification other than "it just happens to be that way".

Yes, I have seen this question and answer. I understand that it's not exactly known when or where these crop up in general, and in fact only a few are generally known. I'm just asking about whether, for this particular case, there is a reason why this fact "should" be the case beyond just the relatively prosaic fact that $6!=8\times 9\times 10$, the missing factors from $7!$.

So my question is: Is it a fact that seems coincidental but actually has a very good rationale, like that $e^{\pi\sqrt{163}}\approx 262537412640768743.999999999999\approx 262537412640768744$, which has to do with Heegner numbers and a bunch of other things I don't quite understand? Or is it a pure coincidence like the fact that $e^{\pi}-\pi\approx 19.999099979\approx 20$?

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    $\begingroup$ This came up recently: math.stackexchange.com/questions/3903361/… $\endgroup$
    – Favst
    Dec 22, 2020 at 18:11
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    $\begingroup$ Do you want more? Consider Fibonacci sequence $$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,\ldots$$ Well, $144$ is the only perfect square in the infinite sequence $\endgroup$
    – Raffaele
    Dec 22, 2020 at 18:13
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    $\begingroup$ Incidentally, $10!$ seconds is exactly $6$ weeks. $\endgroup$
    – Joe
    Dec 22, 2020 at 18:22
  • $\begingroup$ @Favst Fascinating! I'm not generally familiar with group theory, but that is just delightful. $\endgroup$ Dec 22, 2020 at 18:26
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    $\begingroup$ @Raffaele 1 is also a perfect square. $\endgroup$
    – jjagmath
    Aug 13, 2021 at 0:31

1 Answer 1


almost pseudo-Pythagorean-seeming

Check out the ff answer in one of the linked posts

$6!\cdot 7!=10!$. Is there a natural bijection between $S_6\times S_7$ and $S_{10}$?

The answer begins with

It may be connected with, of all things, the 3−4−5 right triangle!


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